You need to use the information the problem provides: `alpha + beta = 90^o` , hence, you need to substitute `90^o` for `alpha + beta ` in product such that:

`sin 2alpha + sin 2beta = 2 sin 90^o*cos (alpha - beta)`

You should substitute 1 for `sin 90^o` such that:

`sin 2alpha + sin 2beta = 2 *cos (alpha - beta)`

Hence, the last line proves the identity `sin 2alpha + sin 2beta = 2 *cos (alpha - beta)` , under the given condition `alpha + beta = 90^o` . The identity as stated in the original question is incorrect.